Problem: A curve is defined by the parametric equations $x=t^2+1$ and $y=t^3$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{4}$ (Choice B) B $\dfrac{3}{2}$ (Choice C) C $\dfrac{3}{2t}$ (Choice D) D $\dfrac{3}{4t}$
Explanation: We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\dfrac{3t}{2}$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\ &=\dfrac{\dfrac{d}{dt}\left(\dfrac{3t}{2}\right)}{\dfrac{d}{dt}(t^2+1)} \\\\ &=\dfrac{\left(\dfrac{3}{2}\right)}{2t} \\\\ &=\dfrac{3}{4t} \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=\dfrac{3}{4t}$.